Question

In the given figure, AB ⊥ BC, GF ⊥ BC and DE ⊥ AC. Prove that ∆ADE ∼ ∆GCE.

Answer 1

We have:

In ΔABC,
$\angle$1 + $\angle$4 = 90°
In ΔGFC,
$\angle$1 + $\angle$2 =90°
Hence, $\angle$1 + $\angle$4 = $\angle$1 + $\angle$2
⇒ $\angle$4 = $\angle$2
⇒ $\angle$A = $\angle$G ..........(i)
Also $\angle$E = $\angle$F ( Each measures 90°) ................(ii)
From (i) and (ii), we get AA similarity for ΔADE and ΔGCF.
⇒ ∆ADE ∼ ∆GCF